888 ENGINEERING THERMODYNAMICS

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\M-therm\Th16-2.pm5

Sol. Pressure in the tank, p 1 = 284 kN/m^2 (gauge)

= 284 + 100 = 384 kN/m^2 (absolute)

Temperature in the tank, T 1 = 24 + 273 = 297 K

Diameter at the outlet of the nozzle, D = 20 mm = 0.02 m

∴ Area, A 2 =

π

4 × 0.02

(^2) = 0.0003141 m 2
R = 287 J/kg K, γ = 1.4
(Two points are considered. Point 1 lies inside the tank and point 2 lies at the exit of the
nozzle).
Maximum flow rate, mmax :
Equation of state is given by p = ρRT or ρ =
p
RT
∴ρ 1 =
p
RT
1
1
= 384 10
287 297
×^3
×
= 4.5 kg/m^3
The fluid parameters in the tank correspond to the stagnation values, and maximum flow
rate of air is given by,
mmax = 0.685 A 2 p 11 ρ ...[Eqn. (16.32)]
= 0.685 × 0.0003141 384 10××^3 4 5. = 0.283 kg/s
Hence maximum flow rate of air = 0.283 kg/s (Ans.)
Example 16.15. A large vessel, fitted with a nozzle, contains air at a pressure of 2500 kN/
m^2 (abs.) and at a temperature of 20°C. If the pressure at the outlet of the nozzle is 1750 kN/m^2 ,
find the velocity of air flowing at the outlet of the nozzle.
Take : R = 287 J/kg K and γ = 1.4.
Sol. Pressure inside the vessel, p 1 = 2500 kN/m^2 (abs.)
Temperature inside the vessel, T 1 = 20 + 273 = 293 K
Pressure at the outlet of the nozzle, p 2 = 1750 kN/m^2 (abs.)
R = 287 J/kg K, γ = 1.4
Velocity of air, V 2 :
V 2 =
2
1
(^11)
1
2
1
1
γ
γρ
γ
γ
−
F
HG
I
KJ
−
F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
−
pp
p
...[Eqn. (16.27)]
where, ρ 1 =
p
RT
1 p RT
1
From equation of state :
ρ

F
HG
I
KJ

2500 10
287 293
×^3
× = 29.73 kg/m
3
Substituting the values in the above equation, we get
V 2 =
214
14 1
2500 10
29 73
1 1750
2500
×^3111
−
F
HG
I
KJ
× × −F
HG
I
KJ
L
N
M
M
M
O
Q
P
P
P
−
.
..
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